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Let $M_t$ be a continuous time real valued martingale, and $\mathcal F_t$ its natural filtration.

Suppose that $\mathcal F_t \setminus \mathcal F_s$ is nonempty for all $t > s$.

Let $\mathcal G$ be a sigma algebra, and define the filtration $\mathcal H_t := F_t \vee \mathcal G$.

Question: Is it true that $M$ is a $\mathcal H_t$ martingale if and only if $\mathcal G$ is independent of $\mathcal F_t$ for all $t$?

Remark: The if direction follows from a monotone class argument.

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  • $\begingroup$ I you wish the condition to be necessary (I think it it not, but for the moment, I have no counterexample), you should at least assume that ALL $\mathcal{F}$-martingales are still $\mathcal{H}$-martingales. This property is called immersion of filtrations. Otherwise, the null martingale provides trivial counterexamples. $\endgroup$
    – Christophe Leuridan
    Commented Jun 29, 2022 at 14:22
  • $\begingroup$ Thanks for your comment - though counterexamples like the null martingale are ruled out by the condition that $\mathcal F_t \setminus \mathcal F_s$ be nontrivial. $\endgroup$
    – Nate River
    Commented Jun 29, 2022 at 15:34
  • $\begingroup$ Ah yes, you assume that $\mathcal{F}$ is the natural filtration of $M$. Anyway, the answer is negative, see the answer below. $\endgroup$
    – Christophe Leuridan
    Commented Jun 29, 2022 at 19:49

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I think that I have a counterexample. Let $(X,Y)$ be a Brownian motion in $\mathbb{R}^2$. Then $M = \int_0^\cdot X_s \mathrm{d}Y_s$ is a martingale, in the natural filtration of $(X,Y)$, in its own filtration $(\mathcal{F}_t)_{t \ge 0}$ and also in $(\mathcal{F}_t \vee \sigma(X))_{t \ge 0}$. Yet, $X$ is not independent of $M$ since $\langle M \rangle = \int_0^\cdot X_s^2\mathrm{d}s$ is not deterministic.

Remark: in this example, $\mathcal{F}$ is not immersed in $\mathcal{H}$ since $X$ is no more a martingale in $\mathcal{H}$.

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    $\begingroup$ Very nice counterexample! $\endgroup$
    – Nate River
    Commented Jun 30, 2022 at 0:10
  • $\begingroup$ @Nate River. Thank you. I still wonder whether your conclusion holds if you assume that that all $\mathcal{F}$-martingales are still $\mathcal{FH$-martingales. The counterexample I gave does not answer this alternative question. $\endgroup$
    – Christophe Leuridan
    Commented Jun 30, 2022 at 6:13

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